How do you convert $(2, \frac{11 π}{6})$ $(2, (11pi)/6)$ from polar to cartesian coordinates?

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How do you convert $(2, \frac{11 π}{6})$ $(2, (11pi)/6)$ from polar to cartesian coordinates?

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Answer 1 · 2017-08-29T21:21:06

$(\sqrt{3}, - 1)$ $(sqrt 3, -1)$

Explanation:

$(2, \frac{11 π}{6})$ $(2, (11pi)/6)$ is in the polar form $(r, θ)$ $(r, theta)$ , and we have to convert it into the Cartesian form $(x, y)$ $(x,y)$ . We can do this by using the following formulas:

$x = r cos θ$ $x = r cos theta$

$y = r sin θ$ $y = r sin theta$

In this case, $r = 2$ $r = 2$ and $θ = \frac{11 π}{6}$ $theta = (11pi)/6$ .

$x = 2 cos (\frac{11 π}{6}) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$ $x = 2 cos ((11pi)/6) = 2 * sqrt3/2 = sqrt3$

$y = 2 sin (\frac{11 π}{6}) = 2 \cdot - \frac{1}{2} = - 1$ $y = 2 sin ((11pi)/6) = 2 * -1/2 = -1$

The Cartesian coordinate is $(\sqrt{3}, - 1)$ $(sqrt 3, -1)$ .

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How do you convert $(2, \frac{11 π}{6})$ $(2, (11pi)/6)$ from polar to cartesian coordinates?

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How do you convert $(2, \frac{11 π}{6})$ $(2, (11pi)/6)$ from polar to cartesian coordinates?